You can rewrite the expression as
$$y=3^{\cos (2 t)+2}+2^{3 \sin (t)+4 \cos (t)}$$
The first piece will reach a maximum value $(3^3=27)$ when $t=0$ and the second will reach a maximum value $(2^5=32)$ when $t=2 \cot ^{-1}(3)$. This means that the maximum value of $y$ will not exceed $27+32=59$; on the other side, the maximum value of $y$ will be larger than $3^3+2^4=43$.
Plotting the function, you could see that the maximum occurs "not very far" of $t=0$. Using composition of Taylor series (this is quite tedious),
$$y=43+48 \log (2)\,t+ \left(72 \log ^2(2)-32 \log (2)-54 \log (3)\right)t^2+
O\left(t^3\right)$$
Computing the derivative and making it equal to zero gives
$$t_{max}=-\frac{12 \log (2)}{36 \log ^2(2)-16 \log (2)-27 \log (3)}\approx 0.354603$$ which gives $y_{max}\approx 48.4354$.
A numerical optimization would lead to $t_{max}\approx 0.330073$ and $y_{max}\approx 48.4634$.
For sure, you could use Newton method to find the zero of
$$y'=\log (2) 2^{3 \sin (t)+4 \cos (t)} (3 \cos (t)-4 \sin (t))-2 \log (3) \sin (2 t)
3^{\cos (2 t)+2}$$ and the iterates would be
$$\left(
\begin{array}{cc}
n & t_n \\
0 & 0.000000 \\
1 & 0.354603 \\
2 & 0.329806 \\
3 & 0.330073
\end{array}
\right)$$
Finally, being very lazy, using a small programmable pocket calculator set in degrees, using steps of $1 ^{\circ}$, the maximum of $y$ was reached for $t=19 ^{\circ}=0.331613$ radians and found to be $48.4633$.
